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Chapter 17: Problem 22

(a) Integrate \(\frac{1}{x\left(1-x^{2}\right)}\) with respect to \(x\). (b) Find the mean value of \(\sin ^{5} x\) over the range \(x=0\) to \(x=\pi / 2\). (c) Obtain an approximate value of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sqrt{\sin \theta} \mathrm{d} \theta\) by the trapezium rule using five ordinates. Work as accurately as your tables will allow.

Short Answer

Expert verified
The integral of \( \frac{1}{x(1-x^2)} \) is \( \frac{1}{2} \ln\left| \frac{x(1+x)}{1-x} \right| + C \). The mean value of \( \sin^5 x \) from \( 0 \) to \( \frac{\pi}{2} \) is \( \frac{8}{15} \). The approximate value using the trapezium rule is approximately 0.693.

Step by step solution

01

Step 1a: Set up the integral

Consider the given integral: \[ \int \frac{1}{x(1-x^{2})} \, dx \]
02

Step 2a: Simplify the integrand using partial fractions

Rewrite the integral using partial fractions. Suppose: \[ \frac{1}{x(1-x^{2})} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} \]Combine the fractions on the right-hand side and equate coefficients to solve for \(A\), \(B\), and \(C\).
03

Step 3a: Solve for partial fraction constants

After finding the constants, the integrand becomes: \[ \frac{1}{x(1-x^{2})} = \frac{1}{2x} + \frac{1}{2(1-x)} - \frac{1}{2(1+x)} \]Now, integrate term by term.
04

Step 4a: Integrate each term

The integral can now be written as: \[ \frac{1}{2} \int \frac{1}{x} \, dx + \frac{1}{2} \int \frac{1}{1-x} \, dx - \frac{1}{2} \int \frac{1}{1+x} \, dx \]Integrate each term separately.
05

Step 5a: Combine the results

The integrals become: \[ \frac{1}{2} \ln|x| - \frac{1}{2} \ln|1-x| + \frac{1}{2} \ln|1+x| + C \]Combine them into a single logarithm: \[ \frac{1}{2} \ln\left| \frac{x(1+x)}{1-x} \right| + C \]
06

Step 1b: Set up the mean value formula

The mean value of a function \(f(x)\) over the interval \([a,b]\) is given by: \[ \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \]For our function, \(f(x) = \sin^5 x\), and the interval \([0, \frac{\pi}{2}]\).
07

Step 2b: Integrate the function \(\sin^5 x\)

Find the integral: \[ \int_{0}^{\frac{\pi}{2}} \sin^5 x \, dx \]This can be solved using reduction formulas or integration techniques.
08

Step 3b: Calculate the mean value

Once the integral is evaluated, divide the result by the length of the interval \(\frac{\pi}{2} - 0 = \frac{\pi}{2}\) to find the mean value.
09

Step 1c: Define the trapezium rule

The trapezium rule for approximating an integral \(\int_{a}^{b} f(x) \, dx\) using \(n\) ordinates is: \[ \frac{h}{2} \left[ f(a) + 2 \sum_{i=1}^{n-1} f(a + ih) + f(b) \right] \]where \(h=\frac{b-a}{n-1}\).
10

Step 2c: Calculate \(h\) and the ordinates

For \(n=5\) ordinates, and limits \(a=\frac{\pi}{6}\) and \(b=\frac{\pi}{2}\): \[ h = \frac{\frac{\pi}{2} - \frac{\pi}{6}}{4} = \frac{\pi}{6} \]
11

Step 3c: Evaluate the function at each ordinate

Determine the value of \(\sqrt{ \sin \theta }\) at each ordinate: \(\theta = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{\pi}{2}\)
12

Step 4c: Apply the trapezium rule formula

Using the values from the previous step, apply the trapezium rule formula to approximate the integral. Sum these values and multiply by \(\frac{h}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

partial fraction decomposition
Partial fraction decomposition is a method used to break down a complex fraction into simpler fractions that are easier to integrate.
Imagine a fraction like \(\frac{1}{x(1-x^2)} \), which is challenging to integrate directly.
By expressing it as a sum of simpler fractions, such as \(\frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} \), we can integrate each term separately.
  • Find constants: Solve for \A, B, \ and \C \ by equating coefficients.
  • Simplify the integral: Once you have the constants, the integrand becomes much simpler.
  • Integrate each term: Now, integrate the resulting simpler fractions term by term.
For our example, after finding the constants, it is decomposed into: \(\frac{1}{x(1-x^2)} = \frac{1}{2x} + \frac{1}{2(1-x)} - \frac{1}{2(1+x)} \) leading to straightforward integration.
definite integrals
Definite integrals are used to find the exact area under a curve over a specific interval.
For instance, the definite integral of \(\frac{1}{x(1-x^2)} \) from \0 \ to \1\ gives the area bounded by the curve and the x-axis in that range.
  • Setup: Identify the function and the interval.
  • Integrate: Use an antiderivative or integration techniques to find the integral of the function.
  • Evaluate: Compute the difference between the values of the antiderivative at the upper and lower limits of the interval.
In our example, to integrate \(\frac{1}{x(1-x^2)} \), we need to first decompose it into partial fractions, then integrate and finally evaluate at the limits.
trapezium rule
The trapezium rule, also known as the trapezoidal rule, helps approximate definite integrals.
It breaks the area under a curve into trapezoids, calculates their areas, and sums them.
  • Formula: Use \(\frac{h}{2} [f(a) + 2 \sum_{i=1}^{n-1} f(a + ih) + f(b)] \) where \h \ is the width of each segment.
  • Calculate \h\: \[ h = \frac{b-a}{n-1} \]
  • Evaluate function: Find the function's values at specified points called ordinates.
For instance, to approximate \(\theta = \frac{\frac{\text{pi}}{6}}{\frac{\text{pi}}{2}} = \frac{\text{pi}}{6} \), we split the interval into five segments, find \sqrt{ \sin} \theta \ at each ordinate, and sum their areas. This approximates the integral's value.
mean value theorem for integrals
The mean value theorem for integrals states that for a continuous function \(f(x)\) on the interval \([a,b]\), there exists a point \(c\) such that: \[ f(c) = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx \] This is the idea of finding the average value of a function over an interval.
  • Identify the function and interval.
  • Integrate: Calculate the definite integral of the function over the interval.
  • Average: Divide the integral by the length of the interval (\text\((b-a)\)).
For example, to find the mean value of \(\text{sin} \^5 {x} \) from \(\frac{0}{\frac{\text{pi}}{2}}\), first calculate \[ \int_{0}^{\frac{\text{pi}}{2}} \text {sin} \^5 {x} d{x}\],then divide the result by \(\frac{\text {pi}}{2} \ - 0 = \ \frac{\text{pi}}{2} \), to find the average value.

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Most popular questions from this chapter

Show that the curves whose equations are $$ y-1=x^{3} \quad \text { and } \quad y+3=3 x^{2} $$ intersect at a point on the \(x\)-axis, and find the coordinates of this point. Show also that the only other point at which the curves meet is \((2,9)\) and that the curves have a common tangent there. Sketch the two curves on the same diagram. Show that the area of the finite region bounded by the curves is \(27 / 4\), and find the \(x\) coordinate of the centroid of this region.

\(R\) is the region in the first quadrant bounded by the \(y\)-axis, the \(x\)-axis from 0 to \(\frac{1}{2} \pi\), the line \(x=\frac{1}{2} \pi\) and part of the curve \(y=(1+\sin x)^{\frac{1}{2}}\). (a) Show that, when \(R\) is rotated about the \(x\)-axis through four right angles, the volume of the solid formed is \(\frac{1}{2} \pi(\pi+2)\). (b) Use the trapezium rule with three ordinates to show that the area of \(R\) is approximately \(0.63 \pi\).

The table below shows the values of \(y\) obtained experimentally for the given values of \(x\). Show graphically that, allowing for small errors of observation, there is a relation of the form \(y-2=k(1+x)^{n} \quad\) and find approximate values of \(k\) and \(n\). \(\begin{array}{cccccc}x & 4 & 8 & 15 & 19 & 24 \\ y & 2.45 & 2.60 & 2.80 & 2.89 & 3.00\end{array}\)

Find the \(x\) coordinate of the turning point of the curve whose equation is $$ y=\frac{a}{x}+\ln x $$ where \(x>0\) and \(a>0, \quad\) and determine whether this turning point is a maximum or a minimum. Deduce the range of values of the constant \(a\) for which \(y \geqslant 0\) for all \(x>0\) In the case when \(a=1, \quad\) find the area and the \(x\) coordinate of the centroid of the region bounded by the curve, the \(x\)-axis and the ordinates \(x=1\) and \(x=2 . \quad\) Express both answers in terms of \(\ln 2\).

A curve joining the points \((0,1)\) and \((0,-1)\) is represented parametrically by the equations $$ x=\sin \theta, \quad y=(1+\sin \theta) \cos \theta $$ where \(0 \leqslant \theta \leqslant \pi\). Find \(\mathrm{d} y / \mathrm{d} x\) in terms of \(\theta\), and determine the \(x, y\) coordinates of the points on the curve at which the tangents are parallel to the \(x\)-axis and of the point at which the tangent is perpendicular to the \(x\)-axis. Sketch the curve. The region in the quadrant \(x \geqslant 0, \quad y \geqslant 0\) bounded by the curve and the coordinate axes is rotated about the \(x\)-axis through an angle of \(2 \pi\). Show that the volume swept out is given by $$ V=\pi \int_{0}^{1}(1+x)^{2}\left(1-x^{2}\right) d x $$ Evaluate \(V\), leaving your result in terms of \(\pi\).
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